Two identical pucks are on an air table. Puck A has an initialvelocity of 2.3 m/

Question

Two identical pucks are on an air table. Puck A has an initialvelocity of 2.3 m/s in the positivex-direction. Puck B is at rest. Puck A collideselastically with puck B and A moves off at 1.0 m/s at an angle of +60° above thex-axis. What is the speed and direction of puck B afterthe collision? (Let angles above the +x-axis to bepositive and below to be negative.)
What is the magnitude ___ m/s What is the direction ___ degrees Two identical pucks are on an air table. Puck A has an initialvelocity of 2.3 m/s in the positivex-direction. Puck B is at rest. Puck A collideselastically with puck B and A moves off at 1.0 m/s at an angle of +60° above thex-axis. What is the speed and direction of puck B afterthe collision? (Let angles above the +x-axis to bepositive and below to be negative.)
What is the magnitude ___ m/s What is the direction ___ degrees

Explanation / Answer

Sice mometum is conserved in general m1V = m1V1 +m2V2 In particular we must consider the x a dn y components in x direction and since masses are the same mV = mV1 cos (60) + mV2cos() V = V1 cos (60) + V2cos() 2.3 =  1.0/2 + V2cos() 1.8 = V2cos() in y-direction we have 0= mV1 sin (60) +mV2sin() in y-direction we have 0= mV1 sin (60) +mV2sin() 3 / 2 = - V2sin() 0.866 = - V2sin() _________ Now we have two equations and two unknowns 1.8 = V2cos() and 0.866= - V2sin() sinceV2 = 1.8/cos() then 0.866 = - [1.8 ]sin() /cos() 0.866 = -  1.8 tan() = arcTan( -0.866 /1.8 ) = -25.7 deg and now since we know that V2 = 1.8/cos() we have V2 = 1.8/cos(-25.7) V2 = 1.99 m/s or V2 = 2.0 m/s

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